Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.*;

public class Main
{
	public static void main(String args[])
	{
		Scanner in=new Scanner(System.in);
		int n=in.nextInt();
		double p[]=new double[n+1];
		double dp[][]=new double[n+1][n+1];
		for(int i=1;i<=n;i++)
			p[i]=in.nextDouble();
		for(int i=0;i<=n;i++)
			dp[0][i]=0;
		dp[1][0]=1-p[1];
		for(int i=2;i<=n;i++)
			dp[i][0]=dp[i-1][0]*(1-p[i]);
		dp[1][1]=p[1];
		for(int i=2;i<=n;i++)
			dp[i][i]=dp[i-1][i-1]*(p[i]);
		for(int i=2;i<=n;i++)
		{
			for(int j=1;j<i;j++)
				dp[i][j]=dp[i-1][j]*(1-p[i])+dp[i-1][j-1]*p[i];
		}
		double ans=0;
		for(int i=n/2+1;i<=n;i++)
			ans+=dp[n][i];
		System.out.println(ans);
	}
}